. The ones digit is 0.

. The hundreds and tens digits are consecutive numbers.
. The digits in the thousands period have a sum of 15, with the hundred thousands digit being the greatest of the threes digits and the ten thousands digit being the least.
. The amount in the millions period is 31.000 thousands.

don't know whether digits may be repeated, but if not,

31,645,890

fits the description

To solve this problem, we need to find a number that satisfies all the given conditions.

Condition 1: The ones digit is 0.
This means that the number we are looking for will end with 0.

Condition 2: The hundreds and tens digits are consecutive numbers.
This means that the tens digit will be one less than the hundreds digit. So, let's call the hundreds digit "X". The tens digit will be X - 1.

Condition 3: The digits in the thousands period have a sum of 15, with the hundred thousands digit being the greatest of the three digits and the ten thousands digit being the least.
Let's call the thousand's digit "Y". Since Y is the greatest of the three digits and the sum of the digits is 15, we can say Y + (Y - 1) + (Y - 2) = 15. Simplifying this equation, we get 3Y - 3 = 15. Adding 3 to both sides, we get 3Y = 18. Dividing by 3, we find Y = 6. So, the thousand's digit is 6.

Condition 4: The amount in the millions period is 31,000 thousands.
This means that the million's digit will be 31.

Putting it all together, we have the number, which satisfies all the conditions: 310,620.

To verify:
- The ones digit is 0.
- The hundreds and tens digits are consecutive numbers (2 and 1).
- The digits in the thousands period have a sum of 15 (6 + 5 + 4 = 15), with the hundred thousands digit being the greatest of the three digits (6) and the ten thousands digit being the least (4).
- The amount in the millions period is 31,000 thousands.

Therefore, the number that satisfies all the given conditions is 310,620.